To solve this problem, we will use Ampere's circuital law to determine the magnetic field inside and outside a long straight wire of circular cross-section carrying a steady current.$\\$Ampere's circuital law states:$\\ \oint B \cdot dl = \mu_0 I_{enc} \\$where $B$ is the magnetic field, $dl$ is an infinitesimal element of the loop, $\mu_0$ is the permeability of free space, and $I_{enc}$ is the current enclosed by the loop.$\\$Consider a wire of radius $R$ carrying a total current $I. \\$1. Inside the wire ($r < R$):$\\$The current density $J$ is uniform, so $J = \frac{I}{\pi R^2}. \\$The current enclosed by a loop of radius $r$ is $I_{enc} = J \cdot \pi r^2 = \frac{I}{\pi R^2} \cdot \pi r^2 = \frac{I r^2}{R^2}. \\$Applying Ampere's law:$\\ B \cdot 2 \pi r = \mu_0 \cdot \frac{I r^2}{R^2} \\ B = \frac{\mu_0 I r}{2 \pi R^2} \\$This shows that the magnetic field inside the wire is a linearly increasing function of $r. \\$2. Outside the wire ($r \geq R$):$\\$The entire current $I$ is enclosed by the loop.$\\$Applying Ampere's law:$\\ B \cdot 2 \pi r = \mu_0 I \\ B = \frac{\mu_0 I}{2 \pi r} \\$This shows that the magnetic field outside the wire decreases with $\frac{1}{r}. \\$Therefore, the correct option is:$\\$Option 3: A linearly increasing function of distance $r$ up to the boundary of the wire and then decreasing one with $\frac{1}{r}$ dependence for the outside region.